including continued fraction

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$1+\displaystyle       \frac{1}{
\displaystyle 1 + \frac{1}{
\displaystyle 1 + \frac{1}{
\displaystyle 1 + \frac{1}{
              1 + \cdots}}}}
=\frac{1+\sqrt{5}}{2}$
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$p+\displaystyle       \frac{1}{
\displaystyle 3p+ \frac{1}{
\displaystyle 5p+ \frac{1}{
\displaystyle 7p+ \frac{1}{
              9p+ \cdots}}}}
=\frac{e^{\frac{2}{p}}+1}{e^{\frac{2}{p}}-1}$ (J.H.Lambert)
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$\displaystyle       \frac{x}{
\displaystyle 1 - \frac{x}{
\displaystyle 3 - \frac{x}{
\displaystyle 5 - \frac{x}{
                      7 - \cdots}}}}
=\tan{x}$
 
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$\displaystyle       \frac{x}{
\displaystyle 1 - \frac{\frac{x^2}{1\cdot3}}{
\displaystyle 1 - \frac{\frac{x^2}{3\cdot5}}{
\displaystyle 1 - \frac{\frac{x^2}{5\cdot7}}{
                      1 - \cdots}}}}
=\tan{x}$
 
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$\displaystyle       \frac{e^{-\frac{2\pi}{5}}}{
\displaystyle 1 + \frac{e^{-2\pi}}{
\displaystyle 1 + \frac{e^{-4\pi}}{
\displaystyle 1 + \frac{e^{-6\pi}}{
                     1 + \cdots}}}}
=\sqrt{\frac{5+\sqrt{5}}{2}}-\frac{1+\sqrt{5}}{2}$
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$\displaystyle     \frac{e^{-\frac{\sqrt{3}\pi}{5}}}{
\displaystyle 1 - \frac{e^{-\sqrt{3}\pi}}{
\displaystyle 1 + \frac{e^{-2\sqrt{3}\pi}}{
\displaystyle 1 - \frac{e^{-3\sqrt{3}\pi}}{
              1 + \cdots}}}}
=\frac{\sqrt{30+6\sqrt{5}}-3-\sqrt{5}}{4}$
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$\displaystyle       \frac{\sqrt{2}e^{-\frac{2\pi}{8}}}{
\displaystyle 1- \frac{e^{-2\pi}}{
\displaystyle 1+e^{-2\pi}- \frac{e^{-4\pi}}{
\displaystyle 1+e^{-4\pi}- \frac{e^{-6\pi}}{
              1+e^{-6\pi}- \cdots}}}}=
\sqrt{\sqrt{2}-1}$
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$\displaystyle     \frac{e^{-\frac{2\pi}{3}}}{
\displaystyle 1 - \frac{e^{-2\pi}+e^{-4\pi}}{
\displaystyle 1 + \frac{e^{-4\pi}+e^{-8\pi}}{
\displaystyle 1 - \frac{e^{-6\pi}+e^{-12\pi}}{
              1 + \cdots}}}}
=\frac{\sqrt{6\sqrt{3}}-1-\sqrt{3}}{4}$
----------
$2+\displaystyle       \frac{1}{
\displaystyle 1 + \frac{1}{
\displaystyle 2 + \frac{2}{
\displaystyle 3 + \frac{3}{
              4 + \cdots}}}}
=e$

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$1+\displaystyle       \frac{2}{
\displaystyle 3 + \frac{4}{
\displaystyle 5 + \frac{6}{
\displaystyle 7 + \frac{8}{
              9 + \cdots}}}}
=\frac{1}{\sqrt{e}-1}$
----------
$\displaystyle       \frac{1}{
\displaystyle n-1 + \frac{1}{
\displaystyle 1 + \frac{1}{
\displaystyle 3n-2 + \frac{1}{
\displaystyle 1 + \frac{1}{
              5n-2 + \cdots}}}}}
=\tan{(\frac{1}{n})$
----------
$\displaystyle     \frac{q^{\frac{1}{5}}}{
\displaystyle 1 + \frac{q}{
\displaystyle 1 + \frac{q^2}{
\displaystyle 1 + \frac{q^3}{
              1 + \cdots}}}}
=q^{\frac{1}{5}} \cdot \frac{\displaystyle\prod_{j=0}^{\infty}(1-q^{5j+1})\displaystyle\prod_{j=0}^{\infty}(1-q^{5j+4})}{\displaystyle\prod_{j=0}^{\infty}(1-q^{5j+2})\displaystyle\prod_{j=0}^{\infty}(1-q^{5j+3})}
~~~~~~~~~~(|q|<1)$
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$\displaystyle     \frac{1}{
\displaystyle 1 + \frac{q+q^2}{
\displaystyle 1 + \frac{q^2+q^4}{
\displaystyle 1 + \frac{q^3+q^6}{
              1 + \cdots}}}}
=\frac{\displaystyle\prod_{j=0}^{\infty}(1-q^{6j+5})\displaystyle\prod_{j=0}^{\infty}(1-q^{6j+1})}{\displaystyle\prod_{j=0}^{\infty}(1-q^{6j+3})^2}
~~~~~~~~~~(|q|<1)$
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$\displaystyle     \frac{1}{
\displaystyle 1 + \frac{q}{
\displaystyle 1 +q+ \frac{q^2}{
\displaystyle 1 +q^2+ \frac{q^3}{
              1 +q^3+ \cdots}}}}
=\frac{\displaystyle\prod_{j=0}^{\infty}(1-q^{2j+1})}{\displaystyle\prod_{j=0}^{\infty}(1-q^{4j+2})^2}
~~~~~~~~~~(|q|<1)$
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$2+\displaystyle                     \frac{2\cdot1^2+1}{
\displaystyle 2\cdot1^2+5\cdot1+2 - \frac{2\cdot2^2+2}{
\displaystyle 2\cdot2^2+5\cdot2+2 - \frac{2\cdot3^2+3}{
\displaystyle 2\cdot3^2+5\cdot3+2 - \frac{2\cdot4^2+4}{
              2\cdot4^2+5\cdot4+2 - \cdots}}}}
=\frac{1}{\frac{\sqrt{2}}{\sin{\sqrt{2}}}-1}$
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$\displaystyle       \frac{1}{
\displaystyle x+1- \frac{1^2}{
\displaystyle x+3- \frac{2^2}{
\displaystyle x+5- \frac{3^2}{
              x+7- \cdots}}}}=
\int_{0}^{\infty}\frac{e^{-z}}{x+z}dz$
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$\sum_{n=1}^{k-1} \frac{2n}{(n^2+r^2)^2}+
\displaystyle       \frac{1}{
\displaystyle (k-\frac{1}{2})^2+\frac{1+4r^2}{4} + \frac{\frac{1^4(1^2+4r^2)}{4(2\cdot1-1)(2\cdot1+1)}}{
\displaystyle (k-\frac{1}{2})^2+\frac{1}{4}(2\cdot1^2+2\cdot1+1+4r^2) - \frac{\frac{2^4(2^2+4r^2)}{4(2\cdot2-1)(2\cdot2+1)}}{
\displaystyle (k-\frac{1}{2})^2+\frac{1}{4}(2\cdot2^2+2\cdot2+1+4r^2) - \frac{\frac{3^4(3^2+4r^2)}{4(2\cdot3-1)(2\cdot3+1)}}{
                     (k-\frac{1}{2})^2+\frac{1}{4}(2\cdot3^2+2\cdot3+1+4r^2) - \cdots}}}}
=\frac{1}{r}\int_{0}^{\infty}\frac{x}{e^x-1}\sin{(rx)}dx~~~~~~~~~~x>\frac{1}{2}~~~~~r>0$